   Chapter 30, Problem 58PE

Chapter
Section
Textbook Problem

Integrated ConceptsIn a laboratory experiment designed to duplicate Thomson's determination of q e / m e , a beam of electrons having a velocity of 6.00 × 10 7   m / s enters a 5.00 × 10 − 3   T magnetic field. The beam moves perpendicular to the field in a path having a 6.80-cm radius of curvature. Determine q e / m e from these observations, and compare the result with the known value.

To determine

qeme from observations and then compare the result with the known value of qeme

Explanation

Given Data:

In a laboratory experiment designed to duplicate Thomson's experiment of qeme , a beam of electrons having a velocity of 6×107m/s enters a 5.00×103T magnetic field. The beam moves perpendicular to the field in a path having a 6.8cm radius of curvature

Formula Used:

Centripetal force is calculated as

F=mv2r

Where F= Force on electrons in a magnetic field

m= Mass of electrons

v= Velocity of electons

Also, magnetic force is calculated as

Fmag=qvB

Where B= Magnetic field

v= Velocity of electron

q= Charge of electron

Calculation:

In Thomson's experiment, centripetal force is balanced by magnetic force.

Thus, we have

qvB=mv2r

qm=v2rvB

qm=vrB

We have B=5

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